Can a particle in an inverse square potential $$V(r)=-1/r^{2}$$ in $d=3$ spatial dimensions be solved exactly? Also please explain me the physical significance of this potential in comparison with Coulomb potential? That problem was talking about positive repulsive potential and what I am looking for is an attractive potential.

  • 1
    $\begingroup$ This question (v6) deserves to be reopened. The currently marked duplicate instead asks about any inverse power law potential, and only contains a heuristic discussion of the inverse square potential case. $\endgroup$
    – Qmechanic
    Jul 17, 2019 at 8:56

2 Answers 2


One may show that the inverse square potential

$$V(r)~=~\frac{\hbar^2}{2m}\frac{c}{r^2}~\propto~\frac{1}{r^2}, \tag{1}$$

in $d$ spatial dimensions mathematically speaking has three alternatives, depending on the dimensionless proportionality constant $$c_d~\equiv~\frac{(d-1)(d-3)}{4}+c~\equiv~\varkappa^2-\frac{1}{4}~\in~\mathbb{R}.\tag{2}$$

  1. Case $c_d < -\frac{1}{4}$: The spectrum is unbounded from below, i.e. the system is unstable.

  2. Case $c_d\geq \frac{3}{4}$: There are no bound states at all.

  3. Case $-\frac{1}{4} \leq c_d < \frac{3}{4}$: It is possible to define asymptotic boundary conditions (ABCs) at $r=0$ / selfadjoint extensions of the Hamiltonian, such that the spectrum is bounded from below. Some of these extensions have bound states, others have not.

From a physics point of view, it may seem unnatural that conclusions depend on adjustable ABCs imposed at $r=0$. The usual physics lore is that an inverse square potential is only an effective description, and that presumably new physics should emerge to resolve the singularity at $r=0$. From now on in this answer, we only discuss the purely mathematical problem at hand, even if it is somewhat academic.

First of all, angular excitations only push the energy up, never down, so it is enough to analyze spherically symmetric $s$-waves $$ \psi(r) ~\equiv~r^{\frac{1-d}{2}}u(r) , \qquad u(r)~\equiv~\sqrt{r}~ v(r) ,\tag{3}$$ and hence the energy functional simplifies to

$$\begin{align}\frac{2m}{\hbar^2} \frac{\langle \psi | \hat{H} \psi \rangle}{{\rm Vol}(S^{d-1})} &~=~ \int_{\mathbb{R}_+} \!r^{d-1}~ dr~ \psi(r)^{\ast} \left(- \frac{\partial^2}{\partial r^2}-\frac{d-1}{r}\frac{\partial}{\partial r} +\frac{c}{r^2}\right) \psi(r) \cr &~=~ \int_{\mathbb{R}_+} \! dr~ u(r)^{\ast} \left(- \frac{\partial^2}{\partial r^2}+\frac{c_d}{r^2}\right) u(r)\cr &~=~ \int_{\mathbb{R}_+} \! dr~ v(r)^{\ast} \left(- \frac{\partial}{\partial r}r\frac{\partial}{\partial r}+\frac{\varkappa^2}{r}\right) v(r). \tag{4}\end{align} $$

1) Sketched proof of case $c_d< -\frac{1}{4}$: Let us study a trial/test function

$$ u(r)~=~ r^{p}e^{-r/2}, \qquad \frac{1}{2}~<p~<~\sqrt{-c_d}, \tag{5} $$

in the limit

$$p ~\searrow~ \frac{1}{2}. \tag{6}$$

The square norm

$$\frac{||\psi||^2}{{\rm Vol}(S^{d-1})}~=~ \int_{\mathbb{R}_+} \! dr~ |u(r)|^2~=~ (2p)! ~\searrow~ 1 \quad\text{for}\quad p ~\searrow~ \frac{1}{2} \tag{7}$$

is finite. The energy functional (4) simplifies to (after integration by parts)

$$\begin{align}\frac{2m}{\hbar^2} \frac{\langle \psi | \hat{H}| \psi \rangle}{{\rm Vol}(S^{d-1})} &~\stackrel{(4)}{=}~ \int_{\mathbb{R}_+} \! dr \left( |u^{\prime}(r)|^2 +\frac{c_d}{r^2} |u(r)|^2\right) \cr &~=~ \int_{\mathbb{R}_+} \! dr \left( \left(p-\frac{r}{2}\right)^2+c_d \right)r^{2p-2}e^{-r}\cr &~=~ \underbrace{(p^2+c_d)}_{<0}\underbrace{\int_{\mathbb{R}_+} \! dr ~r^{2p-2}e^{-r}}_{=(2p-2)!}+\text{finite terms}\cr &~\to~ -\infty\quad\text{for}\quad p ~\searrow~ \frac{1}{2},\tag{8} \end{align}$$

and is unbounded from below. $\Box$

In light of the previous section, let us assume that $\varkappa^2~\equiv~c_d+\frac{1}{4}\geq 0$ from now on. The radial TISE

$$\frac{\hbar^2}{2m}\left(-\frac{1}{r} \frac{\partial}{\partial r}r\frac{\partial}{\partial r}+\frac{\varkappa^2}{r^2}\right) v(r) ~=~ -|E|v(r), \qquad v(r\!=\!\infty)~=~0, \tag{9}$$

for bound states becomes the modified Bessel equation. The solution

$$ \begin{align}v(r)~~\propto~&~ K_{\varkappa}(\rho) ~~\propto~~ -\frac{\sin (\varkappa\pi)}{\pi/2}K_{\varkappa}(\rho) ~=~I_{\varkappa}(\rho)-I_{-\varkappa}(\rho) \cr ~~\stackrel{\rho\ll 1}{\sim}&~~ \frac{(\rho/2)^{\varkappa}}{\Gamma(1+\varkappa)} -\frac{(2/\rho)^{\varkappa}}{\Gamma(1-\varkappa)} , \end{align} \tag{10}$$ $$ \rho ~\equiv~\frac{\sqrt{2m|E|}}{\hbar}r, \tag{11}$$

is a modified Bessel function of the second kind. The problem (9) has a scale symmetry, so that the energy spectrum becomes unbounded from below unless we impose a pertinent ABC at $r=0$.

2) Sketched proof of case $c_d\geq \frac{3}{4}\Leftrightarrow \varkappa\geq 1$: The $\psi$-wavefunction (10) is not square integrable at $r=0$. In other words, pertinent ABC at $r=0$ prohibits the $\psi$-wavefunction (10). $\Box$

3) Sketched proof of case $-\frac{1}{4} \leq c_d < \frac{3}{4}\Leftrightarrow 0\leq \varkappa<1$: We now impose ABC (12) at $r=0$:

$$ v(r)~~\propto~~(k_0r)^{\varkappa} + \lambda (k_0r)^{-\varkappa} + O(r), \qquad k_0r ~\ll~1, \qquad \lambda~\in~\mathbb{R}, \tag{12}$$



is the Compton wavenumber, i.e. the reciprocal reduced Compton wavelength. Here $\lambda\in\mathbb{R}$ is a fixed dimensionless parameter. Comparing eqs. (10) & (11), in the case $0< \varkappa<1$, this leads to a single bound state

$$ E~=~- 2mc^2 \left|\lambda\frac{\Gamma(1-\varkappa)}{\Gamma(1+\varkappa)} \right|^{-1/\varkappa} \quad \text{for}\quad \lambda~<~0,\tag{14}$$

and no bound states if $\lambda\geq 0$. The case $\varkappa=0$ has similar conclusions. See Ref. 3 for details. $\Box$


  1. A.M. Essin & D.J. Griffiths, Quantum mechanics of the $1/x^2$ potential, Am. J. Phys. 74 (2006) 109.

  2. L.D. Landau & E.M. Lifshitz, QM, Vol. 3, 3rd ed, 1981; $\S$35.

  3. D.M. Gitman, I.V. Tyutin & B.L. Voronov, arXiv:0903.5277. (Hat tip: JamalS.)


This is the Calogero potential. It seems at least physically sensible. Note the inequality,

$$\int_{\mathbb R^3} \frac{1}{4r^2}|\psi(x)|^2 \mathrm d^3x \leq \int_{\mathbb R^3}|\nabla\psi(x)|^2 \mathrm d^3x$$

which holds for any $\psi \in C^\infty_0(\mathbb R^3)$. With a potential $V=cr^{-2}$ the Hamiltonian is something like $H = -\nabla^2 + cr^{-2}$ and we have that,

$$\langle H \rangle = \int_{\mathbb R^3} \bar \psi(x)\left(-\nabla^2 + cr^{-2}\right)\psi(x)\, \mathrm d^3x = \int_{\mathbb R^3} |\nabla \psi(x)|^2 + cr^{-2}|\psi(x)|^2 \mathrm d^3x.$$

For a suitably chosen potential, namely with $-c \leq \frac14$, we have that $\langle H \rangle \geq 0$, or in words, we are guaranteed to have the Hamiltonian bounded from below, which is physically sensible. This also allows us to have some choice of self-adjoint extension, via the Friedrichs extension.


For $c \geq \frac34$, there exists only one self-adjoint Hamiltonian, whose spectrum is $\mathbb R_+$. The normalised generalised eigenfunctions are,

$$u_{1,E}(x) = \sqrt{\frac{x}{2}} J_\varkappa(\sqrt{E}x)$$

for $E \geq 0$ and a particular Bessel function, $J_\varkappa$.

For $-\frac14 < c < \frac34$, there is a one-parameter $U(1)$-family of self-adjoint Hamiltonians. The normalised eigenfunctions are,

$$u_{2,\lambda,E}(x) = \sqrt{\frac{1}{x}}\frac{J_\varkappa(\sqrt E x) + \gamma\sqrt{x} J_{-\varkappa}(\sqrt E x)}{\sqrt{1+2\gamma \cos\pi\varkappa} + \gamma^2}$$

where $\gamma = \lambda \frac{\Gamma(1-\varkappa)}{\Gamma(1+\varkappa)}(E/4k^2_0)^\varkappa$. For $\lambda \geq 0$ we have the same spectrum, but for negative $\lambda$, one has,

$$\mathrm{spec} \, H_{2,\lambda} = \left\{-4k^2_0 \bigg\rvert \lambda \frac{\Gamma(1-\varkappa)}{\Gamma(1+\varkappa)}\bigg\rvert^{-1/\varkappa} \right\} \cup [0,\infty)$$

which has a bound state. You can find more in Self-adjoint extensions and spectral analysis in Calogero problem which is the most self-contained, complete treatment I could find; the other cases are treated, as well as additional aspects of the problem.

  • $\begingroup$ Doesn't this potential have no bound states at all ? I tried solving for this potential and I am not getting any discrete bound states solutions. I have gone through the following paper arxiv.org/abs/quant-ph/0202091 In page number six just below equation (32) , the author says as follows"The spectrum becomes continuous and unbounded from below despite its appearance, the hamiltonian is not self adjoint" What does it mean can some one please explain? $\endgroup$
    – user135580
    Dec 1, 2017 at 13:29
  • $\begingroup$ Can someone answer my last quwstions $\endgroup$
    – user135580
    Dec 10, 2017 at 10:27

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